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The long exact cohomology sequence

Suppose \(\mathfrak{a}\) and \(\mathfrak{b}\) are \(\mathfrak{g}\)-modules,
where the representation is denoted by \(d\) in both cases.

Given \(\alpha\in Hom_{\mathfrak{g}}(\mathfrak{a},\mathfrak{b})\) (this means that \(\alpha d_1(x)=d_1(x)\alpha \)
for all \(x\in\mathfrak{g}\)),
one extends \(\alpha\) to a linear map
from \(C^n(\mathfrak{g},\mathfrak{a})\) to \(C^n(\mathfrak{g},\mathfrak{b})\) by
\[(\alpha^n a_n)(x_1,\cdots,x_n)=
\alpha a_n(x_1,\cdots,x_n)\]
for \(a_n\in C^n(\mathfrak{g},\mathfrak{a})\ .\)

It follows that \(\alpha^{\cdot}\) leaves cocycles and coboundaries invariant and induces a map
\[[\alpha^n]:H^n(\mathfrak{g},\mathfrak{a})\rightarrow H^n(\mathfrak{g},\mathfrak{b})\ .\]
One writes \([\alpha^{\cdot}]\) for this family of maps.

Notice that the elements in \(H_\mathfrak{g}(\mathfrak{g},\cdot)\) are just the \(\mathfrak{g}\)-invariant elements
in the \(\mathfrak{g}\)-module, and equivalence classes are to be identified with their representing elements, since there is nothing to divide out since \(d^{-1}=0\ .\)

Thus the sequence is exact at \(H_\mathfrak{g}(\mathfrak{g},\mathfrak{b})\ .\)

remark

The map \([\beta]\) is not necessarily surjective. For example, suppose that \(\mathfrak{g}\) is onedimensional,
with basiselement \(x\) acting on \(\mathfrak{b}=\mathbb{C}^2\) by
\[ d_1(x)e_1=0\]
\[d_1(x)e_2=e_1\]

Take \(\mathfrak{a}=\mathbb{C} e_1\ .\) Then \(H_\mathfrak{g}(\mathfrak{g},\mathfrak{b})=\mathbb{C} e_1\) maps to
\(0\) in \(\mathfrak{c}=\mathfrak{b}/\mathfrak{a}\ ,\) but \(H_\mathfrak{g}(\mathfrak{g},\mathfrak{c})=\mathbb{C}e_2+\mathfrak{a}\)
is nonzero.\(\square\)

One measures the lack of surjectivity of \([\beta]\) by constructing a connecting homomorphism that embeds the left exact sequence (2) into a long exact sequence of cohomology spaces as follows.

Next we turn to the proof of the exactness of the cohomological sequence.
It follows directly from the definition of
\([\delta^n]\) that
\[ \mathrm{im}\ [\beta^n]\subset \ker [\delta^n],\quad \mathrm{im} [\delta^n] \subset \ker [\alpha^{n+1}].\]
For indeed, take \([c_n]=[\beta^n][b_n]\) with \( d^n b_n =0\ .\)

One defines \([\delta^n]\) by constructing \(a_{n+1}\) by \(\alpha^{n+1} a_{n+1} =d^n b_n\)
but this is zero.

Since \(\alpha^{n+1}\) is injective, this means that \(a_{n+1}=0,\) or, in other words,
that \( [\delta^n][\beta^n][b_n]=[0].\)

exact sequence maps

Let \(\mathfrak{a}\subset\mathfrak{b}\) and \(\tilde{\mathfrak{a}}\subset\tilde{\mathfrak{b}}\)
be \(\mathfrak{g}\)-modules, and \( f\in Hom_{\mathfrak{g}}(\mathfrak{b},\tilde{\mathfrak{b}})\)
such that \(f(\mathfrak{a})\subset\tilde{\mathfrak{a}}\ .\)
Let \(\mathfrak{c}=\mathfrak{b}/\mathfrak{a}\) and \(\tilde{\mathfrak{c}}=\tilde{\mathfrak{b}}/\tilde{\mathfrak{a}}\ .\)
Then \(f\) induces maps \(\underline{f}\in Hom_{\mathfrak{g}}(\mathfrak{a},\tilde{\mathfrak{a}})\)
and \(\overline{f}\in Hom_{\mathfrak{g}}(\mathfrak{c},\tilde{\mathfrak{c}})\) by restriction and passing to the quotient, respectively.